【记忆搜索法】FatMouse and Cheese

国际惯例Po主叨一叨，今天做题又暴露出了许多问题，果然Po还是个小菜鸡啊...

首先抛开题目不谈，只谈记忆搜索法，dp，说到底，这个dp数组到底是用来干什么的？

 里面存的应该是通过dfs算出的从这个点出发能得到的最大值！

 只是第一次dfs(x,y)都要递归很多次才能得到从(x,y)这个点出能获得的最大值，如果把这个值存到dp里，下一次dfs只需要访问dp的值就行，不用再递归！

做这个题的时候，一开始我用循环去找这个矩阵里的最大值，超时了一遍又一遍，这很明显就是对dp数组的意义理解得不够到位。

OK,上题目上代码:

FatMouse has stored some cheese in a city. The city can be considered as a square grid of dimension n: each grid location is labelled (p,q) where 0 <= p < n and 0 <= q < n. At each grid location Fatmouse has hid between 0 and 100 blocks of cheese in a hole. Now he's going to enjoy his favorite food.

FatMouse begins by standing at location (0,0). He eats up the cheese where he stands and then runs either horizontally or vertically to another location. The problem is that there is a super Cat named Top Killer sitting near his hole, so each time he can run at most k locations to get into the hole before being caught by Top Killer. What is worse -- after eating up the cheese at one location, FatMouse gets fatter. So in order to gain enough energy for his next run, he has to run to a location which have more blocks of cheese than those that were at the current hole.

Given n, k, and the number of blocks of cheese at each grid location, compute the maximum amount of cheese FatMouse can eat before being unable to move.

Input Specification

There are several test cases. Each test case consists of

• a line containing two integers between 1 and 100: n and k
• n lines, each with n numbers: the first line contains the number of blocks of cheese at locations (0,0) (0,1) ... (0,n-1); the next line contains the number of blocks of cheese at locations (1,0), (1,1), ... (1,n-1), and so on.

The input ends with a pair of -1's.

Output Specification

For each test case output in a line the single integer giving the number of blocks of cheese collected.

Sample Input

3 1

1 2 5

10 11 6

12 12 7

-1 -1

Output for Sample Input

37

-------------------------

#include <cstdio>#include <iostream>#include <algorithm>#include <string.h>using namespace std;int n,k;int dp[105][105];int maps[105][105];int MAX=0;int dir[4][2]={1,0,0,1,-1,0,0,-1};int max(int x,int y){if(x>y)return x;elsereturn y;}int dfs(int x,int y){int tx,ty;int i,j;int MAXX=0;int t;if(dp[x][y]) return dp[x][y];for(i=0;i<4;i++){for(j=1;j<=k;j++){tx=x+dir[i][0]*j;//因为只能走一个方向ty=y+dir[i][1]*j;if(tx>=0 && tx<n && ty>=0 && ty<n){if(maps[tx][ty]>maps[x][y]){t=dfs(tx,ty);//找到值;MAXX=max(MAXX,t);//把较大的值保存在MAXX里}}}}dp[x][y]=MAXX+maps[x][y];//最大的MAXX值和当前的maps里的值相加才是dp应该保存的值return dp[x][y];}int main(){int i,j;while(~scanf("%d %d",&n,&k)){if(n==-1 && k==-1)break;memset(dp,0,sizeof(dp));for(i=0;i<n;i++){for(j=0;j<n;j++){scanf("%d",&maps[i][j]);}}printf("%d\n",dfs(0,0));}return 0;}